Industrial Engineering Journal
ISSN: 0970-2555
Volume : 53, Issue 7, No.3, July : 2024
GENERALIZED SEMI PRE CONTINUOUS MAPPING IN NEUTROSOPHIC TOPOLOGICAL SPACES
Chandramathi N, Department of Mathematics, Government Arts College, Udumalpet-642126, Tamil Nādu, India.E-mail: drmathimaths@gmail.com
Rajeshwaran N, Research Scholar, Department of Mathematics, Government Arts College, Udumalpet-642126, Tamilnadu, India.
Abstract: In this manuscript, we inaugurate Neutrosophic generalized semi pre continuous mapping, Neutrosophic generalized semi pre irresolute mapping and Neutrosophic generalized semi pre compact space. We investigate its properties. Also, we add some improvisation of neutrosophic generalized semi pre continuous mapping.
Keywords: Neutrosophic generalized semi pre closed sets, Neutrosophic generalized semi pre continuous mapping, Neutrosophic generalized semi pre irresolute mapping , Neutrosophic generalized semi pre compact space.
1. Introduction
In 2014, Salama, Smarandache and Valeri [9] initiate idea of Neutrosophic closed sets and Neutrosophic continuous functions. Salama, Alblowi [11] launch the conceptualization of generalized Neutrosophic set and generalized Neutrosophic topological spaces. Wadel and Smarandache [14] popularized NOS. Ishwarya and Bageerathi [8] introduced the view of NSO sets in Neutrosophic topological spaces. Rajeshwaran N and Chandramathi N [25] introduced the Neutrosophic generalized semi pre closed sets in Neutrosophic topological spaces This manuscript we establish Neutrosophic generalized semi pre continuous mapping.. We study their concepts of Neutrosophic generalized semi pre continuous mapping.
2. Preliminaries
Definition 2.1: [10] A neutrosophic topology (NT for short) a non-empty set X is a family τN of neutrosophic subsets in X satisfying the following axioms (NT1)0N,1N ∈τN (NT2)G1 ∩G2 ∈τN (NT3)∪Gi ∈τN ,∀{Gi:i∈J}⊆τN
Here (X1 ,τN) is called a neutrosophic topological space (NTS for short).
Definition 2.2: [10] Let A1������A2 be two Neutrosophic Sets (NS for Short) of the form
A1 ={⟨X,μA1(X),σA1(X),γA1(X)⟩:xϵX} , A2 ={⟨X,μA2(X),σA2(X),γA2(X)⟩:xϵX} .
(a) A1 ⊆A2 if and only if μA1(X)≤μA2(X),σA1(X)≤σA2(X)andγA1(X)≥γA2(X)forallx∈X
(b) A1 C ={⟨X,γA1(X),1 σA1(X),μA1(X)⟩:xϵX}
(c)A1 ∩A2 ={⟨X,μA1(X)⋀μA2(X),σA1(X)⋀σA2(X),γA1(X)⋁γA2(X)⟩:xϵX}
(d) A1 ∪A2 ={⟨X,μA1(X)⋁μA2(X),σA1(X)⋁σA2(X),γA1(X)⋀γA2(X)⟩:xϵX}
We can use the symbol A1 ={⟨X,μA(X),σA(X),γA(X)⟩:xϵX}
The Neutrosophic Sets define by 0N ={〈X,0,0,1〉∶x∊X} and 1N ={〈X,1,1,0〉∶x∊X}.
Definition 2.3: [10] Let (X,τ) be an NTS and A=〈x,μ��,σA,γ��〉 be an NTS in X. Then the neutrosophic interior and an neutrosophic closure are defined by int(A)=∪{G/GisaNOSinXandG⊆A} . cl(A)=∩{K/KisaNCSinXandA⊆K}
Industrial Engineering Journal
ISSN: 0970-2555
Volume : 53, Issue 7, No.3, July : 2024
Definition 2.4: [25] Let (Ҳ,τƝ) be a neutrosophic topological space. A subset Ą of (Ҳ,τƝ) is called Neutrosophic generalized semi pre closed [ƝƓŚƤ -closed] set if spclƝ (Ą) ⊆Մ , whenever Ą⊆Մ and Մ is Neutrosophic open set
Complement of Neutrosophic generalized semi pre closed set is called the Neutrosophic generalized semi pre [ƝƓŚƤ -open] open set.
Definition 2.5:[26] A Neutrosophic topological space (Ҳ,τƝ) is said to be ƝƓŚƤ normal if for any pair of disjoint ƝƓŚƤ closed sets Ą and Ƀ, there exist disjoint Neutrosophic open sets M and N such that Ą ⊂Մ,Ƀ⊂Ṿ
Preposition 2.5: [13] Let (X1 ,τN) , (X2,σN) be two neutrosophic topological spaces, if g:(X1 ,τN) →(X2,σN) neutrosophic continuous then it is N- continuous
Definition 2.6: [13] A neutrosophic topological space (X1 ,τN) is said to be neutrosophic T1/2 if any Neutrosophic closed set in (X1 ,τN) is neutrosophic closed in (X1 ,τN)
Definition 2.7: [22] A mapping g:(X1 ,τN) →(X2,σN) is defined Neutrosophic semi continuous mapping if g -1 (K) is NSOS in (X1 ,τN) for each NOS K (X2,σN)
3. Neutrosophic Generalized Semi Pre-Continuous mapping
In this section, we Introduce Neutrosophic generalized semi pre continuous mapping and investigate some of its properties.
Definition 3.1.1: A mapping φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) is called Neutrosophic generalized semi pre continuous (ƝƓŚƤ continuous for short) mapping if φƝ 1(Ɗ) is ƝƓŚƤƇŚ in (Ҳ,τƝ) for every ƝƇŚ Ɗ of (Ƴ,σƝ).
Example 3.1.2: Let Ҳ={a,b,c}, Ƴ={u,v,w} and Ⱪ1 ={x,〈0.2,0.4,0.5〉,〈0.4,0.5,0.5〉,〈0.2,0.5,0.2〉}, Ⱪ2 = {y,〈0.3,0.2,0.3〉,〈0.4,0.5,0.5〉,〈0.4,0.4,0.5〉}
Then τƝ ={0Ɲ,1Ɲ,Ⱪ1} and σƝ ={0Ɲ,1Ɲ,Ⱪ2} are NTS on (Ҳ,τƝ) and (Ƴ,σƝ) respectively.
Define a mapping φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) by φƝ(a)=u,φƝ(b)=v,φƝ(c)=w. Here the neutrosophic set Ⱪ2 �� ={y,〈03,08,03〉,〈05,05,04〉,〈05,06,04〉} is a neutrosophic closed set in (Ƴ,σƝ). Then φƝ 1(Ⱪ2 ��)={x,〈0.3,0.8,0.3〉,〈0.5,0.5,0.4〉,〈0.5,0.6,0.4〉} is ƝƓŚƤƇŚ in (Ҳ,τƝ) . Hence φƝ is ƝƓŚƤ continuous mapping.
Theorem 3.1.3: Every Neutrosophic continuous mapping is a NGSP continuous mapping but not conversely.
Proof: Let φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) be Neutrosophic continuous mapping. Let Ɗ be a ƝƇŚ in (Ƴ,σƝ) Then φƝ 1(Ɗ) is an ƝƇŚ in (Ҳ,τƝ). Since every ƝƇŚ is a ƝƓŚƤƇŚ, φƝ 1(Ɗ) is a ƝƓŚƤƇŚ in (Ҳ,τƝ). Hence φƝ is a ƝƓŚƤ continuous mapping
Example 3.1.4: Let Ҳ={a,b}, Ƴ={u,v} and and Ⱪ1 ={x,〈0.6,0.3,0.2〉,〈0.5,0.2,0.3〉}, Ⱪ2 = {y,〈05,04,03〉,〈04,02,03〉} Then τƝ ={0Ɲ,1Ɲ,Ⱪ1} and σƝ ={0Ɲ,1Ɲ,Ⱪ2} are NTS on (Ҳ,τƝ) and (Ƴ,σƝ) respectively. Define a mapping φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) by φƝ(a)=u,φƝ(b)=v. Here the neutrosophic set Ⱪ2 �� ={y,〈0.3,0.6,0.5〉,〈0.3,0.8,0.4〉} is a neutrosophic closed set in (Ƴ,σƝ). Then φƝ 1(Ⱪ2 ��)={x,〈0.3,0.6,0.5〉,〈0.3,0.8,0.4〉} is ƝƓŚƤƇŚ in (Ҳ,τƝ). Hence φƝ is a ƝƓŚƤ continuous mapping. But φƝ is not neutrosophic continuous mapping since Ⱪ2 �� is neutrosophic closed set in (Ƴ,σƝ) but φƝ 1(Ⱪ2 ��) is not a neutrosophic closed set in (Ҳ,τƝ) as clƝ(φƝ 1(Ⱪ2 ��))=1Ɲ ≠φƝ 1(Ⱪ2 ��).
Theorem 3.1.5: Every ƝƓ continuous mapping is a ƝƓŚƤ continuous mapping but not conversely.
Industrial Engineering Journal
ISSN: 0970-2555
Volume : 53, Issue 7, No.3, July : 2024
Proof: Let φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) be a ƝƓ continuous mapping. Let Ɗ be a ƝƇŚ in (Ƴ,σƝ). Then φƝ 1(Ɗ) is a ƝƓƇŚ in (Ҳ,τƝ). Since every ƝƓƇŚ is a ƝƓŚƤƇŚ, φƝ 1(Ɗ) is a ƝƓŚƤƇŚ in (Ҳ,τƝ). Hence φƝ is a ƝƓŚƤ continuous mapping.
Example 3.1.6: Let Ҳ={a,b} , Ƴ={u,v} and Ⱪ1 ={x,〈0.4,0.4,0.6〉,〈0.5,0.4,0.5〉}, Ⱪ2 ={��,〈0.7,0.6,0.2〉〈0.5,0.6,0.3〉}. Then τƝ ={0Ɲ,1Ɲ,Ⱪ1} and σƝ ={0Ɲ,1Ɲ,Ⱪ2} are ƝƬŚ on (Ҳ,τƝ) and (Ƴ,σƝ) respectively. Define a mapping φƝ ∶ (Ҳ,τƝ) →(Ƴ,σƝ) by φƝ(a)=u,φƝ(b)=v . Here the neutrosophic set Ⱪ2 �� = {��,〈02,04,07〉〈03,04,05〉} is a neutrosophic closed set in (Ƴ,σƝ) . Then φƝ 1(Ⱪ2 ��)= {��,〈0.2,0.4,0.7〉〈0.3,0.4,0.5〉} is ƝƓŚƤƇŚ in (Ҳ,τƝ). Hence φƝ is a ƝƓŚƤ continuous mapping. But φƝ is not neutrosophic generalized continuous mapping since Ⱪ2 �� is neutrosophic closed set in (Ƴ,σƝ) but φƝ 1(Ⱪ2 ��) is not a ƝƓƇŚ in (Ҳ,τƝ) as clƝ (φƝ 1(Ⱪ2 ��))=Ⱪ1 �� .
Theorem 3.1.7: Every ƝŚ continuous mapping is a ƝƓŚƤ continuous mapping but not conversely.
Proof: Let φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) be a ƝŚ continuous mapping. Let Ɗ be a ƝƇŚ in (Ƴ,σƝ). Then φƝ 1(Ɗ) is a ƝŚƇŚ in (Ҳ,τƝ) . Since every ƝŚƇŚ is a ƝƓŚƤƇŚ, φƝ 1(Ɗ) is a ƝƓŚƤƇŚ in(Ҳ,τƝ) . Hence φƝ is a ƝƓŚƤ continuous mapping.
Example 3.1.8: Let Ҳ ={a,b,c} , Ƴ={u,v,w} and Ⱪ1 ={x,〈0.2,0.4,0.6〉,〈0.1,0.7,0.9〉,〈0.3,0.6,0.9〉}, Ⱪ2 ={y,〈0.1,0.2,0.1〉,〈0.4,0.7,0.6〉,〈0.3,0.7,0.9〉}. Then τƝ ={0Ɲ,1Ɲ,Ⱪ1} and σƝ ={0Ɲ,1Ɲ,Ⱪ2} are ƝƬŚ on (Ҳ,τƝ) and (Ƴ,σƝ) respectively. Define a mapping φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) by φƝ(a)=u,φƝ(b)=v,φƝ(c)=w . Here the neutrosophic set Ⱪ2 �� = {��,〈0.1,0.8,0.1〉〈0.6,0.3,0.4〉〈0.9,0.3,0.3〉} is a neutrosophic closed set in (Ƴ,σƝ) . Then
φƝ 1(Ⱪ2 ��)={��,〈01,08,01〉〈06,03,04〉〈09,03,03〉} is ƝƓŚƤƇŚ in (Ҳ,τƝ) Hence φƝ is a ƝƓŚƤ continuous mapping. But φƝ is not neutrosophic generalized continuous mapping Then φƝ is ƝƓŚƤ continuous mapping but not a ƝŚ continuous mapping. Since Ⱪ2 �� is a ƝƇŚ in (Y,σƝ) and
φƝ 1(Ⱪ2 �� ) is not a ƝŚƇŚ in (Ҳ,τƝ) Since intƝ(clƝ(φƝ 1(Ⱪ2 �� )))=1Ɲ ⊈ φƝ 1(Ⱪ2 �� ).
Theorem 3.1.9: Every ƝƤ continuous mapping is a ƝƓŚƤ continuous mapping but not conversely. Proof: Let φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) be a ƝƤ continuous mapping. Let Ɗ be a ƝƇŚ in (Ƴ,σƝ). Then φƝ 1(Ɗ) is a ƝƤƇŚ in(Ҳ,τƝ). Since every ƝƤƇŚ is a ƝƓŚƤƇŚ, φƝ 1(Ɗ) is a ƝƓŚƤƇŚ in (Ҳ,τƝ). Hence φƝ is a ƝƓŚƤ continuous mapping.
Example: 3.1.10: Let Ҳ={a,b} ,Ƴ={u,v} and Ⱪ1 ={x,〈0.2,0.3,0.8〉,〈0.5,0.3,0.5〉},Ⱪ2 ={y,〈0.3,0.5,0.6〉,〈0.2,0.4,0.7〉}. Then τƝ ={0Ɲ,1Ɲ,Ⱪ1} and σƝ ={0Ɲ,1Ɲ,Ⱪ2} are ƝƬ on (Ҳ,τƝ) and (Ƴ,σƝ) respectively Define a mapping φƝ ∶ (Ҳ,τƝ) →(Ƴ,σƝ) by φƝ(a)=u,φƝ(b)=v . Here the neutrosophic set Ⱪ2 �� = {��,〈0.6,0.5,0.3〉〈0.7,0.6,0.2〉} is a neutrosophic closed set in (Ƴ,σƝ) . Then φƝ 1(Ⱪ2 ��)= {��,〈0.6,0.5,0.3〉〈0.7,0.6,0.2〉} is ƝƓŚƤƇŚ in (Ҳ,τƝ).Then φƝ is ƝƓŚƤ continuous mapping but not ƝƤ continuous mapping since Ⱪ2 �� is ƝƇŚ in (Ƴ,σƝ) and φƝ 1(Ⱪ2 �� ) is not a ƝƤƇŚ in (Ҳ,τƝ)
Theorem: 3.1.11: Every ƝŚƤ continuous mapping is a ƝƓŚƤ continuous mapping but not conversely.
Proof: Let φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) be a ƝŚƤ continuous mapping. Let Ɗ be a ƝƇŚ in (Ƴ,σƝ).Then φƝ 1(Ɗ) is a ƝŚƤƇŚ in (Ҳ,τƝ) Since every ƝŚƤƇŚ is ƝƓŚƤƇŚ, φƝ 1(Ɗ) is a ƝƓŚƤƇŚ in (Ҳ,τƝ) Hence φƝ is a ƝƓŚƤ continuous mapping.
Example 3.1.12 : Let Ҳ={a,b} , Ƴ={u,v} and
Industrial Engineering Journal
ISSN: 0970-2555
Volume : 53, Issue 7, No.3, July : 2024
Ⱪ1 ={x,〈0.4,0.4,0.6〉,〈0.4,0.4,0.3〉}, Ⱪ2 ={y,〈0.4,0.5,0.6〉,〈0.3,0.4,0.4〉}. Then τƝ ={0Ɲ,1Ɲ,Ⱪ1} and σƝ ={0Ɲ,1Ɲ,Ⱪ2} are ƝƬŚ on (Ҳ,τƝ) and (Ƴ,σƝ) respectively. Define a mapping φƝ ∶ (Ҳ,τƝ) →(Ƴ,σƝ) by φƝ(a)=u,φƝ(b)=v . Here the neutrosophic set Ⱪ2 �� = {��,〈0.6,0.5,0.4〉〈0.4,0.6,0.3〉} is a neutrosophic closed set in (Ƴ,σƝ) . Then φƝ 1(Ⱪ2 ��)= {��,〈0.6,0.5,0.4〉〈0.4,0.6,0.3〉} is ƝƓŚƤƇŚ in (Ҳ,τƝ). Then φƝ is ƝƓŚƤ continuous mapping, but not ƝŚƤ continuous mapping. Since Ⱪ2 �� is ƝƇŚ in (Ƴ,σƝ) and φƝ 1(Ⱪ2 ��) is not a ƝŚƤƇŚ in (Ҳ,τƝ).
Theorem: 3.1.13: Every Ɲα continuous mapping is a ƝƓŚƤ continuous mapping but not conversely. Proof: Let φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) be a Ɲα continuous mapping. Let Ɗ be a ƝƇŚ in (Ƴ,σƝ) Then φƝ 1(Ɗ) is a ƝαƇŚ in (Ҳ,τƝ) Since every ƝαƇŚ is ƝƓŚƤƇS, φƝ 1(Ɗ) is a ƝƓŚƤƇŚ in (Ҳ,τƝ) Hence φƝ is a ƝƓŚƤ continuous mapping.
Example 3.1.14: Let Ҳ={a,b} , Ƴ={u,v} and Ⱪ1 ={x,〈0.2,0.1,0.7〉,〈0.4,0.4,0.7〉},Ⱪ2 ={y,〈0.9,0.8,0.7〉,〈0.2,0.4,0.6〉}. Then
τƝ ={0Ɲ,1Ɲ,Ⱪ1} and σƝ ={0Ɲ,1Ɲ,Ⱪ2} are ƝƬŚ on (Ҳ,τƝ) and (Ƴ,σƝ) respectively. Define a mapping φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) by φƝ(a)=u,φƝ(b)=v . Here the neutrosophic set Ⱪ2 �� = {��,〈07,02,09〉〈06,06,02〉} is a neutrosophic closed set in (Ƴ,σƝ) . Then φƝ 1(Ⱪ2 ��)= {��,〈0.7,0.2,0.9〉〈0.6,0.6,0.2〉} is ƝƓŚƤƇŚ in (Ҳ,τƝ). Then φƝ is ƝƓŚƤ continuous mapping, but not Ɲα continuous mapping. Since Ⱪ2 �� is ƝƇŚ in (Ƴ,σƝ) but φƝ 1(Ⱪ2 ��) is not a ƝαƇŚ in (Ҳ,τƝ). Theorem: 3.1.15: Let φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) be a mapping where φƝ 1(Ɗ) is a ƝŖƇŚ in (Ҳ,τƝ) for every ƝƇŚ Ɗ in (Ƴ,σƝ). Then φƝ is a ƝƓŚƤ continuous mapping but not conversely. Proof: Assume that φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) is a mapping. Let Ą be a ƝƇŚ in (Ƴ,σƝ). Then
φƝ 1(Ɗ) is a ƝŖƇŚ in (Ҳ,τƝ), by hypothesis. Since every ƝŖƇŚ is ƝƓŚƤƇŚ, φƝ 1(Ɗ) is a ƝƓŚƤƇŚ in (Ҳ,τƝ) HenceφƝ is a ƝƓŚƤ continuous mapping.
Example 3.1.16: Let Ҳ={a,b} , Ƴ={u,v} and Ⱪ1 ={x,〈0.5,0.6,0.5〉,〈0.5,0.4,0.5〉},Ⱪ2 ={y,〈0.5,0.3,0.5〉,〈0.5,0.7,0.5〉}. Then
τƝ ={0Ɲ,1Ɲ,Ⱪ1} and σƝ ={0Ɲ,1Ɲ,Ⱪ2} are ƝƬŚ on (Ҳ,τƝ) and (Ƴ,σƝ) respectively. Define a mapping φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) by φƝ(a)=u,φƝ(b)=v . Here the neutrosophic set Ⱪ2 �� = {��,〈0.5,0.7,0.5〉〈0.5,0.3,0.5〉} is a neutrosophic closed set in (Ƴ,σƝ) . Then φƝ 1(Ⱪ2 ��)= {��,〈0.5,0.7,0.5〉〈0.5,0.3,0.5〉} is ƝƓŚƤƇŚ in (Ҳ,τƝ). Then φƝ is ƝƓŚƤ continuous mapping, but not ƝŖ continuous mapping. Since Ⱪ2 �� is ƝƇŚ in (Ƴ,σƝ) but φƝ 1(Ⱪ2 ��) is not a ƝŖƇŚ in (Ҳ,τƝ)
Theorem: 3.1.17: Let φƝ:(Ҳ1 ,τƝ) →(Ҳ2,σƝ) be a ƝƓŚƤ continuous mapping and let ΨƝ ∶ (Ҳ2,σƝ) →(Ҳ3,ɳƝ) be a Ɲ continuous mapping, then φƝ ₀ΨƝ:(Ҳ1 ,τƝ) →(Ҳ3,ɳƝ) is a ƝƓŚƤ continuous mapping.
Proof: Let Ɗ be a ƝƇŚ in (Ҳ3,ɳN). Then ΨƝ 1(Ɗ) is a ƝƇŚ in (Ҳ2,σƝ), by hypothesis. Since φƝ is a ƝƓŚƤ continuous mapping, φƝ 1(ΨƝ 1(Ɗ)) is a ƝƓŚƤƇŚ in(Ҳ1 ,τƝ). Hence φƝ ₀ΨƝ is a ƝƓŚƤ continuous mapping.
Theorem: 3.1.18: If φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) is a ƝƓŚƤ continuous mapping , then for each ƝƤ(α,β,γ) of (Ҳ,τƝ) and each Ą∊σƝ such that φƝ(ƝƤ(α,β,γ))∊Ą, there exists a ƝƓŚƤOŚ Ƀ of (Ҳ,τƝ) such that ƝƤ(α,β,γ) ∊Ƀ and φƝ(Ƀ)⊆Ą.
Proof: Let ƝƤ(α,β,γ) be a ƝƤ of (Ҳ,τƝ) and Ą∊σƝ such that φƝ(ƝƤ(α,β,γ))∊Ą. Put Ƀ=φƝ 1(Ą). Then by hypothesis, Ƀ is a ƝƓŚƤOŚ in (Ҳ,τƝ) such that ƝƤ(α,β,γ) ∊Ƀ and φƝ(Ƀ)= φƝ(φƝ 1(Ą)))⊆Ą
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ISSN: 0970-2555
Volume : 53, Issue 7, No.3, July : 2024
Theorem: 3.1.19: If φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) is a ƝƓŚƤ continuous mapping, then for each ƝƤ(α,β,γ) of (Ҳ,τƝ) and each Ą∊σƝ such that φƝ(ƝƤ(α,β,γ))qĄ, there exists a ƝƓŚƤOŚ Ƀ of (Ҳ,τƝ) such that ƝƤ(α,β,γ)qɃ and φƝ(Ƀ)⊆Ą.
Proof: Let ƝP(α,β,γ) be a ƝƤ of Ҳ and Ą∊σƝ such that φƝ(ƝƤ(α,β,γ))qĄ. Put Ƀ=φƝ 1(Ą). Then by hypothesis, Ƀ is a ƝƓŚƤƠŚ in (Ҳ,τƝ) such that ƝP(α,β,γ)qɃ and φƝ(Ƀ)=φƝ(φƝ 1(Ą)))⊆Ą
Theorem: 3.1.20: Let φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) is a ƝƓŚƤ continuous mapping. Then φƝ is a ƝŚƤ continuous mapping if (Ҳ,τƝ) is a ƝŚƤƬ1/2 space.
Proof: Let Ɗ be a ƝƇŚ in (Ƴ,σƝ). Then φƝ 1(Ɗ) is a ƝƓŚƤƇŚ in (Ҳ,τƝ), by hypothesis. Since (Ҳ,τƝ) is a ƝŚƤƬ1/2 space, φƝ 1(Ɗ) is a ƝŚƤƇŚ in (Ҳ,τƝ). Hence φƝ is a ƝŚƤ continuous mapping.
Theorem: 3.1.21: Let φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) be a ƝƓŚƤ continuous mapping and let ΨƝ:(Ƴ,σƝ) →(Z,ɳƝ) be a ƝƓ continuous mapping where (Ƴ,σƝ) is a ƝƬ1/2 space. Then φƝ ₀ΨƝ ∶(Ҳ,τƝ) →(Z,ɳƝ) is a ƝƓŚƤ continuous mapping.
Proof: Let Ɗ be a ƝƇŚ in (Z,ɳƝ) Then ΨƝ 1(Ɗ) is a ƝƓƇS in (Ƴ,σƝ), by hypothesis. Since (Ƴ,σƝ) is a ƝƬ1/2 space, ΨƝ 1(Ɗ) is a ƝƇŚ in (Ƴ,σƝ). Therefore φƝ 1(ΨƝ 1(Ɗ)) is a ƝƓŚƤƇŚ in (Ҳ,τƝ), by hypothesis. Hence φƝ ₀ΨƝ is a ƝƓŚƤ continuous mapping.
Theorem: 3.1.22: Let φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) be a mapping from ƝƬ(Ҳ,τƝ) to (Ƴ,σƝ). Then the following conditions are equivalent if (Ҳ,τƝ) and (Ƴ,σƝ) are ƝŚƤƬ1/2 spaces:
(i) φƝ is a ƝƓŚƤ continuous mapping,
(ii) φƝ 1(Ƀ) is a ƝƓŚƤOŚ in (Ҳ,τƝ) for each ƝOŚ Ƀ in (Ƴ,σƝ)
(iii) For every ƝƤp(α,β,γ) in (Ҳ,τƝ) and for every ƝOŚ Ƀ in (Ƴ,σƝ) such that φƝ(ƝƤp(α,β,γ))∈ Ƀ, there exists a ƝƓŚƤOŚ Ą in (Ҳ,τƝ) such that (p(α,β,γ))∈Ą and φƝ(Ą)⊆Ƀ.
Proof: (i)⇔(ii) is obvious, since φƝ 1(ĄC)=(φƝ 1(A))C (ii)⇒(iii) Let Ƀ be a any ƝƠŚ in (Ƴ,σƝ) and let (Ɲp(α,β,γ))∈(Ҳ,τƝ). Given φƝ(Ɲp(α,β,γ))∈Ƀ. By hypothesis φƝ 1(Ƀ) is a ƝƓŚƤOŚ in (Ҳ,τƝ) . Take Ą=φƝ 1(Ƀ) . Now Ɲp(α,β,γ) ∈ φƝ 1(φƝ(Ɲp(α,β,γ))). Therefore φƝ 1(φƝ(Ɲp(α,β,γ)))∈φƝ 1(Ƀ)=Ą. This implies (Ɲp(α,β,γ))∈ Ą and φƝ(Ą)=φƝ(φƝ 1(Ƀ))⊆Ƀ (iii)⇒(i) Let Ą be a ƝƇŚ in (Ƴ,σƝ). Then its complement, say Ƀ=ĄC is a ƝOŚ in (Ƴ,σƝ). Let Ɲp(α,β,γ) ∈(Ҳ,τƝ) and φƝ(Ɲp(α,β,γ))∈Ƀ. Then there exists a ƝƓŚƤOŚ, say Ƈ in (Ҳ,τƝ) such that Ɲp(α,β,γ) ∈Ƈ and φƝ(Ƈ)⊆Ƀ . Now Ƈ⊆φƝ 1(φƝ(Ƈ))⊆φƝ 1(Ƀ) . Thus Ɲp(α,β,γ) ∈φƝ 1(Ƀ) Therefore φƝ 1(Ƀ) is a ƝƓŚƤƠŚ in (Ҳ,τƝ) . That is φƝ 1(ĄC) is a ƝƓŚƤOŚ in (Ҳ,τƝ) and hence φƝ 1(Ą) is a ƝƓŚƤƇŚ in (Ҳ,τƝ). Thus φƝ is a ƝƓŚƤ continuous mapping. Theorem: 3.1.23: Let φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) be a mapping from ƝƬ(Ҳ,τƝ) to ƝƬ(Ƴ,σƝ). Then the following conditions are equivalent if (Ҳ,τƝ) and (Ƴ,σƝ) are ƝŚƤƬ1/2 spaces:
(i) φƝ is a ƝƓŚƤ continuous mapping,
(ii) For each ƝƤp(α,β,γ) in (Ҳ,τƝ) and for every Neutrosophic neighborhood (ƝƝforshort) Ą of φƝ(Ɲp(α,β,γ)), there exists a ƝƓŚƤƠŚ Ƀ in (Ҳ,τƝ) such that (Ɲp(α,β,γ))∈Ƀ⊆φƝ 1(Ą) (iii) For each ƝƤp(α,β,γ) in (Ҳ,τƝ) and for every ƝƝĄ of φƝ(Ɲp(α,β,γ)), there exists a ƝƓŚƤƠŚ Ƀ in (Ҳ,τƝ) such that (Ɲp(α,β,γ))∈Ƀ and φƝ(Ƀ)⊆Ą.
Proof: (i)⇒(ii) Let (Ɲp(α,β,γ))∈(Ҳ,τƝ) and let Ą be a ƝƝ A of φƝ(Ɲp(α,β,γ)). Then there exists a ƝƠŚ Ƈ in (Ƴ,σƝ) such that φƝ(Ɲp(α,β,γ))∈Ƈ⊆Ą. Since φƝ is a ƝƓŚƤ continuous mapping,
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φƝ 1(Ƈ)=Ƀ(say), is a ƝƓŚƤOŚ in (Ҳ,τƝ) and (Ɲp(α,β,γ))∈Ƀ⊆φƝ 1(Ą)
(ii)⇒(iii) Let (Ɲp(α,β,γ))∈(Ҳ,τƝ) and let Ą be a ƝƝ of φƝ(Ɲp(α,β,γ)). Then there exists a ƝƓŚƤOŚ Ƀ in (Ҳ,τƝ) such that (Ɲp(α,β,γ))∈Ƀ⊆φƝ 1(Ą), by hypothesis. Therefore (Ɲp(α,β,γ))∈Ƀ and φƝ(Ƀ)⊆φƝ(φƝ 1(Ą))⊆Ą
(iii)⇒(i) Let Ƀ be a ƝOŚ in (Ƴ,σƝ) and let (Ɲp(α,β,γ))∈φƝ 1(Ƀ). Then φƝ(Ɲp(α,β,γ))∈Ƀ. Therefore Ƀ is a ƝƝ of φƝ(Ɲp(α,β,γ)) Since Ƀ is ƝOŚ, by hypothesis there exists a ƝƓŚƤOŚ Ą in (Ҳ,τƝ) such that (Ɲp(α,β,γ))∈Ą⊆φƝ 1(φƝ(Ą))⊆φƝ 1(Ƀ). Therefore φƝ 1(Ƀ) is a ƝƓŚƤOŚ in (Ҳ,τƝ). Hence φƝ is a ƝƓŚƤ continuous mapping.
Theorem: 3.1.24: Let φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) be a mapping from ƝƬ (Ҳ,τƝ) to ƝƬ (Ƴ,σƝ). Then the following conditions are equivalent if (Ҳ,τƝ) is a ƝŚƤƬ1/2 spaces:
(i) φƝ is a ƝƓŚƤ continuous mapping, (ii) IfɃ is a ƝOŚ in (Ƴ,σƝ) then φƝ 1(Ƀ) is a ƝƓŚƤOŚ in (Ҳ,τƝ), (iii) φƝ 1(intƝ(Ƀ))⊆clƝ(intƝ(clƝ(φƝ 1(Ƀ)))) for every ƝŚ Ƀ in (Ƴ,σƝ) .
Proof: (i)⇔(ii) is obviously true by Theorem 3.1.22
(ii)⇒(iii) Let Ƀ be any ƝŚ in (Ƴ,σƝ). Then intƝ(Ƀ) is a ƝƠŚ in (Ƴ,σƝ). Then φƝ 1(intƝ(Ƀ)) is a ƝƓŚƤOŚ in (Ҳ,τƝ). Since (Ҳ,τƝ) is a ƝŚƤƬ1/2 space, φƝ 1(intƝ(Ƀ)) is a ƝŚƤOŚ in (Ҳ,τƝ). Therefore
φƝ 1(intƝ(Ƀ))⊆clƝ(intƝ(clƝ(φƝ 1(intƝ(Ƀ)))))⊆clƝ(intƝ(clƝ(φƝ 1(Ƀ)))).
(iii)⇒(i) Let Ƀ be a ƝOŚ in (Ƴ,σƝ) . By hypothesis φƝ 1(Ƀ)=φƝ 1(intƝ(Ƀ))⊆ clƝ(intƝ(clƝ(φƝ 1(Ƀ)))). This implies φƝ 1(Ƀ) is ƝβOŚ in (Ҳ,τƝ). Therefore it is a ƝƓŚƤOŚ in (Ҳ,τƝ), and hence φƝ is a ƝƓŚƤ continuous mapping, by Theorem 3.1.22
Theorem: 3.1.25: Let φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) be a mapping from ƝƬ (Ҳ,τƝ) to ƝƬ (Ƴ,σƝ). Then the following conditions are equivalent if (Ҳ,τƝ) and (Ҳ,τƝ) are ƝŚƤƬ1/2 spaces:
(i) φƝ is a ƝƓŚƤ continuous mapping, (ii) intƝ(clƝ(intƝ(φƝ 1(Ƀ)))⊆φƝ 1(spclƝ(Ƀ)) for each ƝƇŚ Ƀ in (Ƴ,σƝ), (iii) φƝ 1(spintƝ(Ƀ))⊆clƝ(intƝ(clƝ(φƝ 1(Ƀ)))) for each ƝOŚ Ƀ of (Ƴ,σƝ), (iv) φƝ(intƝ(clƝ(intƝ((Ą))))⊆clƝ(φƝ(Ą)) for each ƝŚ Ą of (Ҳ,τƝ) .
Proof: (i)⇒(ii) Let Ƀ be a ƝƇŚ in (Ƴ,σƝ). Then φƝ 1(Ƀ) is a ƝƓŚƤƇŚ in (Ҳ,τƝ) . Since (Ҳ,τƝ) is a ƝŚƤƬ1/2 space, φƝ 1(Ƀ) is a ƝŚƤƇŚ. Therefore intƝ(clƝ(intƝ(φƝ 1(Ƀ)))⊆φƝ 1(Ƀ)= φƝ 1(spclƝ(Ƀ)) (ii)⇒(iii) can be easily proved by taking complement in (ii).
(iii)⇒(iv) Let Ą∈(Ҳ,τƝ) . Taking Ƀ=φƝ(Ą) we have Ą⊆φƝ 1(Ƀ). Here intƝ(φƝ(Ą))=
intƝ(Ƀ) is a ƝOŚ in (Ƴ,σƝ) . Then (iii) implies that φƝ 1(spintƝ(intƝ(Ƀ))⊆ clƝ(intƝ(clƝ(φƝ 1(intƝɃ))))⊆clƝ(intƝ(clƝ(φƝ 1(Ƀ)))). Now we have clƝ(intƝ(clƝ(Ą��)))�� ⊆
clƝ(intƝ(clƝ(φƝ 1(Ƀ��)))�� ⊆φƝ 1(spintƝ(intƝ(ɃC)))C . This implies intƝ(clƝ(intƝ((Ą)))⊆
φƝ 1(spclƝ(clƝ(Ƀ))) . Therefore φƝ(intƝ(clƝ(intƝ((Ą))))⊆φƝ(φƝ 1(spclƝ(clƝ(Ƀ))))⊆
clƝ(Ƀ)=clƝ(φƝ(Ą))
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(iv)⇒(i) Let Ƀ be a any ƝƇŚ in (Ƴ,σƝ). Then φƝ 1(Ƀ) is a ƝŚ in (Ҳ,τƝ) . By hypothesis
φƝ(intƝ(clƝ(intƝ((φƝ 1(Ƀ)))))⊆clƝ(φƝ(φƝ 1(Ƀ)))⊆clƝ(Ƀ)=Ƀ. Now
intƝ(clƝ(intƝ(φƝ 1(Ƀ))))⊆φƝ 1(φƝ(intƝ(clƝ(intƝ(φƝ 1(Ƀ))))⊆φƝ 1(Ƀ) . This implies
φƝ 1(Ƀ) is a ƝβƇŚ and hence it is a ƝƓŚƤƇŚ in (Ҳ,τƝ). Thus φƝ is a ƝƓŚƤ continuous mapping.
Theorem: 3.1.26: A mapping φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) is a ƝƓŚƤ continuous mapping if
clƝ(intƝ(clƝ(φƝ 1(Ą))))⊆φƝ 1(clƝ(Ą)) for every ƝŚ Ą in (Ƴ,σƝ)
Proof: Let Ą be a ƝOŚ in (Ƴ,σƝ). Then Ą�� is a ƝƇŚ in (Ƴ,σƝ).Therefore clƝ(Ą��)=Ą��. By hypothesis,
clƝ(intƝ(clƝ(φƝ 1(Ą��))))⊆φƝ 1 (clƝ(Ą��))=φƝ 1(Ą��). Now (intƝ(clƝ(intƝ(φƝ 1(Ą)))))C =
clƝ(intƝ(clƝ(φƝ 1(Ą��))))⊆φƝ 1(Ą��)=(φƝ 1(Ą))C . This implies φƝ 1(Ą)⊆ intƝ(clƝ(intƝ(φƝ 1(Ą)))). Hence φƝ 1(Ą) is a ƝαƠŚ in (Ҳ,τƝ) and hence it is a ƝƓŚƤOŚ in (Ҳ,τƝ). Therefore φƝ is a ƝƓŚƤ continuous mapping, by Theorem 3.1.22. Theorem: 3.1.27: Let φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) be a mapping from ƝƬ (Ҳ,τƝ) to ƝƬ (Ƴ,σƝ). Then the following conditions are equivalent if (Ҳ,τƝ) is a ƝŚƤƬ1/2 spaces: (i) φƝ is a ƝƓŚƤ continuous mapping, (ii) φƝ 1(Ƀ) is a ƝƓŚƤƇŚ in (Ҳ,τƝ) for every ƝƇŚ Ƀ in (Ƴ,σƝ), (iii) intƝ(clƝ(intƝ(φƝ 1(Ą))))⊆φƝ 1(clƝ(Ą)) for every ƝŚ Ą in (Ƴ,σƝ)
Proof: (i)⇔(ii) is obviously true by Definition 3.1.1 (ii)⇒(iii) Let Ą be a ƝŚ in (Ƴ,σƝ). Then clƝ(Ą) is a ƝƇŚ in (Ƴ,σƝ).. By hypothesis, φƝ 1(clƝ(Ą)) is a ƝƓŚƤƇŚ in (Ҳ,τƝ). Since (Ҳ,τƝ) is a ƝŚƤƬ1/2 space, φƝ 1(clƝ(Ą)) is a ƝŚƤƇŚ in (Ҳ,τƝ)
Therefore we have intƝ(clƝ(intƝ(φƝ 1(clƝ(Ą)))))⊆φƝ 1(clƝ(Ą)) . Now intƝ(clƝ(intƝ(φƝ 1(Ą))))⊆intƝ(clƝ(intƝ(φƝ 1(clƝ(Ą)))))⊆φƝ 1(clƝ(Ą)). (iii)⇒(i) Let Ą be a ƝƇŚ in (Ƴ,σƝ). By hypothesis intƝ(clƝ(intƝ(φƝ 1(Ą))))⊆φƝ 1(clƝ(Ą))= φƝ 1(Ą) . This implies φƝ 1(Ą) is a ƝβƇŚ in (Ҳ,τƝ) and hence it is a ƝƓŚƤƇŚ. Thus φƝ is a ƝƓŚƤ continuous mapping.
3.2 NEUTROSOPHIC GENERALIZED SEMI PRE IRRESOLUTE MAPPING
In this part, we endeavor to offer an extensive view of ƝƓŚƤ irresolute mappings in Neutrosophic topological spaces, outlining their theoretical bases, fundamental attributes.
Definition 3.2.1: A map φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) is called a Neutrosophic generalized semi-pre irresolute (ƝƓŚƤ irresolute for short) mapping if φƝ 1(Ɗ) is a ƝƓŚƤ closed sets in (Ҳ,τƝ) for every ƝƓŚƤ closed set Ɗ of (Ƴ,σƝ).
Example 3.2.2: Let Ҳ={a,b} , Ƴ={u,v} and Ⱪ1 ={x,〈05,06,01〉,〈05,04,01〉}, Ⱪ2 ={y,〈05,03,02〉,〈05,07,02〉}. Then τƝ ={0Ɲ,1Ɲ,Ⱪ1} and σƝ ={0Ɲ,1Ɲ,Ⱪ2} are ƝƬŚ on (Ҳ,τƝ) and (Ƴ,σƝ) respectively. Define a mapping φƝ ∶ (Ҳ,τƝ) →(Ƴ,σƝ) by φƝ(a)=u, φƝ(b)=v . Here the neutrosophic set Ⱪ3 = {y,〈0.3,0.3,0.7〉,〈0.2,0.3,0.7〉} is a ƝƓŚƤ closed set in (Ƴ,σƝ) Then φƝ is a ƝƓŚƤ irresolute mapping since φƝ 1(Ⱪ3) is ƝƓŚƤ closed set in (Ҳ,τƝ).
Theorem: 3.2.3: If φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) is a ƝƓŚƤ irresolute mapping then φƝ is a ƝƓŚƤ continuous mapping but not conversely.
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Proof: Let φƝ be a ƝƓŚƤ irresolute mapping. Let Ɗ be any Neutrosophic closed sets in (Ƴ,σƝ) Then Ɗ is a ƝƓŚƤ closed sets and by hypothesis φƝ 1(Ɗ) is a ƝƓŚƤ closed sets in (Ҳ,τƝ). Hence φƝ is a ƝƓŚƤ continuous mapping.
Example 3.2.4: Let Ҳ={a,b} , Ƴ={u,v} and Ⱪ1 ={x,〈0.4,0.3,0.6〉 , 〈0.5,0.3,0.5〉} , Ⱪ2 = {y,〈0.6,0.7,0.3〉,〈0.7,0.8,0.3〉}. Then τƝ ={0Ɲ,1Ɲ,Ⱪ1} and σƝ ={0Ɲ,1Ɲ,Ⱪ2} are ƝƬŚ on (Ҳ,τƝ) and (Ƴ,σƝ) respectively. Define a mapping φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) by φƝ(a)=u, φƝ(b)=v . Then φƝ is ƝƓŚƤ continuous mapping since for a neutrosophic closed set Ⱪ2 �� =
{y,〈0.3,0.3,0.6〉,〈0.3,0.2,0.7〉} in (Ƴ,σƝ), its inverse image φƝ 1(Ⱪ2 ��) is ƝƓŚƤ closed set in (Ҳ,τƝ) . But φƝ is not a ƝƓŚƤ irresolute mapping. For the neutrosophic set Ⱪ3 = {y,〈0.4,0.3,0.6〉, 〈0.5,0.3,0.5〉} is ƝƓŚƤ closed set in (Ƴ,σƝ) but φƝ 1(Ⱪ3) is not a ƝƓŚƤ closed set in (Ҳ,τƝ).
Theorem: 3.2.5: Let φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) and ΨƝ ∶(Ƴ,σƝ)→(Z,ɳƝ) be a ƝƓŚƤ irresolute mapping. Then φƝ oΨƝ ∶(Ҳ,τƝ)→(Z,ɳƝ) is a ƝƓŚƤ irresolute mapping.
Proof: Let Ɗ be a ƝƓŚƤ closed set in (Z,ɳƝ). Then ΨƝ 1(Ɗ) is a ƝƓŚƤ closed set in (Ƴ,σƝ). Since φƝ is a ƝƓŚƤ irresolute, φƝ 1(ΨƝ 1(Ɗ)) is a ƝƓŚƤ closed set in (Ҳ,τƝ), by hypothesis. Hence φƝ oΨƝ is a ƝƓŚƤ irresolute mapping.
Theorem: 3.2.6: Let φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) be a ƝƓŚƤ irresolute mapping and ΨƝ ∶(Ƴ,σƝ)→ (Z,ɳƝ) be a ƝƓŚƤ continuous mapping. Then ΨƝoφƝ ∶(Ҳ,τƝ)→(Z,ɳƝ) is a ƝƓŚƤ continuous mapping.
Proof: Let Ɗ be a Neutrosophic closed set in (Z,ɳƝ). Then ΨƝ 1(Ɗ) is a ƝƓŚƤ closed set in (Ƴ,σƝ). Since φƝ is a ƝƓŚƤ irresolute mapping, φƝ 1(ΨƝ 1(Ɗ)) is a ƝƓŚƤ closed set in (Ҳ,τƝ). Hence φƝ oΨƝ is a ƝƓŚƤ continuous mapping.
Theorem: 3.2.7: Let φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) be a mapping from a ƝƬ (Ҳ,τƝ) into a ƝƬ (Ƴ,σƝ). Then the following conditions are equivalent if (Ҳ,τƝ) and (Ƴ,σƝ) are ƝŚƤƬ1/2 spaces: (i) φƝ is a ƝƓŚƤ irresolute mapping, (ii) φƝ 1(Ɗ) is a ƝƓŚƤ open set in (Ҳ,τƝ) for each ƝƓŚƤ open set in (Ƴ,σƝ), (iii) φƝ 1(spintƝ(Ɗ))⊆spintƝ(φƝ 1(Ɗ)) for each ƝŚ Ɗ of (Ƴ,σƝ), (iv) spclƝ(φƝ 1(Ɗ))⊆φƝ 1(spclƝ(Ɗ)) for each ƝŚ Ɗ of (Ƴ,σƝ).
Proof: (i)⇔(ii) is obvious, since φƝ 1(ɃC)=(φƝ 1(Ƀ))C . (ii)⇒(iii) Let Ɗ be any ƝŚ in (Ƴ,σƝ) and spintƝ(Ɗ)⊆Ɗ. Also φƝ 1(spintƝ(Ɗ))⊆φƝ 1(Ƀ).
Since spintƝ(Ɗ) is a ƝŚƤ open set in (Ƴ,σƝ), it is a ƝƓŚƤ open set in (Ƴ,σƝ). Therefore
φƝ 1(spintƝ(Ɗ)) is a ƝƓŚƤ open set in (Ҳ,τƝ), by hypothesis. Since (Ҳ,τƝ) is a ƝŚƤƬ1/2 space,
φƝ 1(spintƝ(Ɗ)) is a ƝŚƤ open set in (Ҳ,τƝ) . Hence
φƝ 1(spintƝ(Ɗ))=spintƝ(φƝ 1(spintƝ(Ɗ)))⊆spintƝ(φƝ 1(Ɗ))
(iii)⇒(iv) is obvious by taking complement in (iii).
(iv)⇒(i) Let Ɗ be a ƝƓŚƤ closed set in (Ƴ,σƝ). Since (Ƴ,σƝ) is a ƝŚƤƬ1/2 space, Ɗ is a ƝŚƤ closed sets in (Ƴ,σƝ) and spclƝ(Ɗ)=Ɗ. Hence φƝ 1(Ɗ)=φƝ 1(spclƝ(Ɗ))⊇spclƝ(φƝ 1(Ɗ)), by hypothesis. But φƝ 1(Ɗ)⊆spclƝ(φƝ 1(Ɗ)). Therefore spclƝ(φƝ 1(Ɗ))=φƝ 1(Ɗ). This implies φƝ 1(Ɗ) is a ƝŚƤ closed sets and hence it is a ƝƓŚƤ closed set in (Ҳ,τƝ). Thus φƝ is a ƝƓŚƤ irresolute mapping.
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Theorem: 3.2.8: Let φƝ ∶(Ҳ,τƝ) →(Ƴ,σƝ) be a ƝƓŚƤ irresolute mapping from a ƝƬ (Ҳ,τƝ) into (Ƴ,σƝ) . Then φƝ 1(Ɗ)⊆spintƝ(φƝ 1(clƝ(intƝ(clƝ(Ɗ))))) for every ƝƓŚƤ open set Ɗ in (Ƴ,σƝ), if (Ҳ,τƝ) and (Ƴ,σƝ) are ƝŚƤƬ1/2 spaces.
Proof: Let Ɗ be a ƝƓŚƤ open set in (Ƴ,σƝ). Then by hypothesis φƝ 1(Ɗ) is a ƝƓŚƤ open set in (Ҳ,τƝ). Since (Ҳ,τƝ) is ƝŚƤƬ1/2 space, φƝ 1(Ɗ) is a ƝŚƤ open set in (Ҳ,τƝ). Therefore spintƝ(φƝ 1(Ɗ))=φƝ 1(Ɗ). Since (Ƴ,σƝ) is ƝŚƤƬ1/2 space, Ɗ is a ƝŚƤ open set (Ƴ,σƝ) and Ɗ⊆clƝ(intƝ(clƝ(Ɗ))). Now, φƝ 1(Ɗ)= spintƝ(φƝ 1(Ɗ)), implies φƝ 1(Ɗ)⊆spintƝ(φƝ 1(clƝ(intƝ(clƝ(Ɗ)))))
3.3 NEUTROSOPHIC GENERALIZED SEMI PRE COMPACT SPACE.
In this segment, our goal is to furnish a thorough exploration of ƝƓŚƤ compact space within Neutrosophic topological spaces, clarifying their theoretical underpinnings and fundamental characteristics.
Definition: 3.3.1: Let (Ҳ,τƝ) be a ƝTŚ. If a family {〈x,μĢ��(��),σĢ��(x),γĢ��(x)〉:i∈J} of ƝƓŚƤ open sets in (Ҳ,τƝ) satisfies the condition ∪{〈x,μĢ��(��),σĢ��(x),γĢ��(x)〉:i∈J}=1Ɲ, then is called a ƝƓŚƤ open cover of (Ҳ,τƝ).
Definition: 3.3.2: Let (Ҳ,τƝ) be a ƝTŚ. A finite subfamily of a ƝƓŚƤ open cover {〈x,μĢ��(��),σĢ��(x),γĢ��(x)〉:i∈J} of (Ҳ,τƝ), which is also a ƝƓŚƤ open cover of (Ҳ,τƝ) is called a finite subcover of {〈x,μĢ��(��),σĢ��(x),γĢ��(x)〉:i∈J}.
Definition: 3.3.3: A ƝTŚ(Ҳ,τƝ) is called ƝƓŚƤ compact iff every ƝƓŚƤ open cover of (Ҳ,τƝ) has a finite subcover.
Definition: 3.3.4: Let (Ҳ,τƝ) be a ƝTŚ. A family {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉:i∈J} of ƝƓŚƤ closed sets in (Ҳ,τƝ) satisfies the finite intersection property (in short ƝIƤ) iff every finite subfamily {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉:i=1,2,…,n} , of the family satisfies the condition ∩��=1 �� {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉}≠0Ɲ.
Theorem 3.3.5: A ƝTŚ (Ҳ,τƝ) is ƝƓŚƤ compact iff every family {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉:i∈ J} of ƝƓŚƤ closed sets with finite intersection property has a non empty intersection. Proof: Let ƝTŚ (Ҳ,τƝ) is ƝƓŚƤ-compact. Suppose, {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉:i∈J} be any family of ƝƓŚƤ-closed sets in (Ҳ,τƝ) such that ∩{〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉:i∈J}=0Ɲ. Thus, this implies that, {〈x,⋀μⱩ��(��),⋀σⱩ��(x),⋁γⱩ��(x)〉}=0Ɲ, {⋀μⱩ��(��):i∈J}=0, {⋀σⱩ��(��):i∈J}=0 and {⋁γⱩ��(x):i∈J}=1 , ⇒ ∪{〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉:i∈J}=1Ɲ . Thus, {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉:i∈J} is a ƝƓŚƤ open cover of (Ҳ,τƝ). Since (Ҳ,τƝ) is ƝƓŚƤcompact, so every ƝƓŚƤ-open cover of (Ҳ,τƝ) has finite subcover. Therefore, (Ҳ,τƝ) has finite subcover {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉:i=1,2,…,n}. So ∪��=1 �� {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉}=1Ɲ, this implies that {〈x,⋁��=1 �� μⱩ��(��),⋁��=1 �� σⱩ��(x),⋀��=1 �� γⱩ��(x)〉:i∈J}=1Ɲ , ⋁��=1 �� {μⱩ��(��)}=0 , ⋁��=1 �� {σⱩ��(��)}=0 and⋀��=1 �� {γⱩ��(��)}=1 implies that ∩��=1 �� {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉}=0Ɲ, which contradicts to our hypothesis. Hence, every family of ƝƓŚƤ closed set with finite intersection property has a non empty intersection. Conversely, suppose every family of ƝƓŚƤ closed set with finite inersecton property has a non empty intersection. Assume that , {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉:i∈J} is any ƝƓŚƤ open cover of (Ҳ,τƝ), then ∪{〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉:i∈J}=1Ɲ. Therefore {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉:i∈J} is a family of ƝƓŚƤ closed sets in (Ҳ,τƝ) such that ∩{〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉:i∈J}=0Ɲ. By
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Volume : 53, Issue 7, No.3, July : 2024
assumption, we can find a finite subfamily, {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉:i=1,2, ,n} such that ∩��=1 �� {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉}=0Ɲ, which implies, ∪��=1 �� {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉}=1Ɲ. Thus {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉:i=1,2,…,n}=1Ɲ is a finite subcover of (Ҳ,τƝ). Hence (Ҳ,τƝ) is ƝƓŚƤ compact.
Definition: 3.3.6: Let (Ҳ,τƝ) be a ƝTŚ and Ą be a Neutrosophic set in (Ҳ,τƝ). If a family {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉:i∈J} of ƝƓŚƤ open sets in (Ҳ,τƝ) satisfies the condition Ą⊆∪ {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉:i∈J}, then it is called a ƝƓŚƤ open cover of Ą.
Definition: 3.3.7: Let (Ҳ,τƝ) be a ƝTŚ. A finite subfamily of a ƝƓŚƤ open cover {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉:i∈J} of Ą, which is also a ƝƓŚƤ open cover of Ą is called a finite subcover of {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉:i∈J}.
Definition: 3.3.8: The Neutrosophic set Ą=〈x,μĄ,σĄ,γĄ〉 in a ƝTŚ (Ҳ,τƝ) is called ƝƓŚƤ compact iff every ƝƓŚƤ open cover of Ą has a finite subcover.
Theorem 3.3.9: Let (Ҳ,τƝ) be a ƝTŚ. A ƝƓŚƤ closed subset of a ƝƓŚƤ compact space is Neutrosophic compact relative to (Ҳ,τƝ).
Proof: Let Ą be a ƝƓŚƤ closed subset of (Ҳ,τƝ). Let {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉:i∈J} be ƝƓŚƤ open cover of Ą. Then the family {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉:i∈J}∪ĄC is ƝƓŚƤ open cover of (Ҳ,τƝ) . Since (Ҳ,τƝ) is a ƝƓŚƤ compact, there is a finite subfamily {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉:i=1,2, ,n} of ƝƓŚƤ open cover, which also covers (Ҳ,τƝ). If this cover contains ĄC we discard it. Otherwise leave the subcover as it is . Thus, we obtained a finite ƝƓŚƤ open subcover of Ą. So Ą is ƝƓŚƤ compact relative to (Ҳ,τƝ).
Theorem 3.3.10: Let (Ҳ,τƝ) be a ƝTŚ. If (Ҳ,τƝ) is ƝƓŚƤ compact space, then it is compact.
Proof: Suppose, (Ҳ,τƝ) be a ƝƓŚƤ compact. Assume contrary that (Ҳ,τƝ) is not fuzzy compact, then there is atleast one fuzzy open cover {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉:i∈J} of (Ҳ,τƝ) not has a finite subcover, implies that ∪{〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉}=1Ɲ a open cover of (Ҳ,τƝ) such that ∪��=1 �� {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉}=1Ɲ. Since, every Neutrosophic open set is ƝƓŚƤ open set. Therefore a open cover {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉:i∈J} of (Ҳ,τƝ) becomes ƝƓŚƤ open cover of (Ҳ,τƝ) such that ∪��=1 �� {〈x,μⱩ��(��),σⱩ��(x),γⱩ��(x)〉}=1Ɲ, which is a contradiction. Hence, if (Ҳ,τƝ) is ƝƓŚƤ compact space, then it is compact.
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